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This is the second of our many quizzes on Kinematic equations. Click on this to access more such competitive quizzes on Mechanics.

Kinematic equations Quiz 2

physics

Two persons $A$ and $B$ started their straight line motion from the same place. $A$ started with initial velocity 5 m/s and acceleration $3.5m/{{s}^{2}}$ and $B$ started with a initial velocity of 10 m/s and acceleration of $1.5m/{{s}^{2}}$. If the distance travelled $A$ and $B$ in next five second is ${{S}_{1}}$ and ${{S}_{2}}$ then- 

We can also apply the first equation of motion to calculate final velocity at 5 sec than by applying the third equation of motion the required distance can be calculated.  

 $\text{Third equation of motion is given by,}\\ s=ut+\dfrac{1}{2}a{{t}^{2}}\\ \text{Distance travelled by A in 5 sec will be-}\\ {{s}_{1}}=5\times 5+\dfrac{1}{2}\times 3.5\times {{5}^{2}}\\ \Rightarrow{{s}_{1}} =25+43.75\\ \Rightarrow{{s}_{1}} =68.75\\ \text{Distance travelled by B in 5 sec will be-}\\ \Rightarrow {{s}_{2}}=10\times 5+\dfrac{1}{2}\times 1.5\times {{5}^{2}}\\ \Rightarrow {{s}_{2}}=50+18.75\\ \therefore {{s}_{2}}=68.75\\ \text{Distance travelled by both people is equal.}$

Sofia teaches physics to high school kids. One day in a laboratory experiment she throws a ball of mass $m$ with initial velocity $V$ in upward direction by which ball can reach a maximum height of ${{H}_{1}}$. Now she throws the same ball at an angle $\theta ={{60}^{\circ }}$ from the horizontal axis and the ball reaches the height of ${{H}_{2}}$. The value of $\dfrac{{{H}_{1}}}{{{H}_{2}}}$ will be- 

 If the ball is thrown with some angle then its horizontal component of velocity is responsible for horizontal motion and vertical component for upward motion.

  $\text{Third equation of motion will be used to calculate maximum height so,}\\ {{v}^{2}}={{u}^{2}}+2as\\ \text{Maximum height in first condition will be given by,}\\ {{0}^{2}}={{V}^{2}}-2\times g\times {{H}_{1}} \\ \Rightarrow {{H}_{1}}=\dfrac{{{V}^{2}}}{2g}\\ \text{In second condition ball is thrown with an angle of}\,{{60}^{\circ }}\\ \text{with the horizontal to vertical component of velocity is}\, V\sin ({{60}^{\circ }})\\ \text{so maximum height}\,{{H}_{2}}\,\text{will be-}\\ {{H}_{2}}=\dfrac{{{V}^{2}}}{2g}\\ \Rightarrow {{H}_{2}} =\dfrac{{{\left( V\sin ({{60}^{\circ }}) \right)}^{2}}}{2g}\\ \Rightarrow {{H}_{2}} =\dfrac{{{V}^{2}}{{\sin }^{2}}{{60}^{\circ }}}{2g}\\ \Rightarrow {{H}_{2}}=\dfrac{3}{4}\dfrac{{{V}^{2}}}{2g}\\ \Rightarrow \dfrac{{{H}_{1}}}{{{H}_{2}}}=\dfrac{\dfrac{{{V}^{2}}}{2g}}{\dfrac{3}{4}\dfrac{{{V}^{2}}}{2g}}\\ \therefore\dfrac{{{H}_{1}}}{{{H}_{2}}}=\dfrac{4}{3}$

A horse is moving randomly in a horizontal direction. Rahul draws a velocity time graph of its straight line motion as given in figure. Now he wants to calculate the acceleration of horse at t=8 sec and that will be equal to – <img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/7835ac51-ef17-4ce1-9daf-840346f59fc63508265132078752764.png"/>

In a velocity-time graph slope determines the acceleration of a body. 

 $\text{Velocity at t=8 sec is 15m/s}\\ \text{Velocity at t= 10 sec is 5 m/s}\\ \text{Acceleration will be equal to,}\\ \Rightarrow a=\dfrac{\Delta v}{\Delta t}\\ \Rightarrow a=\dfrac{5-15}{10-8} \\ \therefore a=-5m/s$

 Ajay loves to play with gun toys which makes sound after hitting their target. One day he randomly fired his toy gun with initial velocity $v$ and heard a sound of a bullet hitting the wall after some time. Now he is calculating the total time between the bullet fire and hearing the sound from the wall. The total time will be equal to-$[\text{Speed of sound is V and distance of Ajay and wall is S}]$ 

 The total time taken will be a summation of time taken by ball in travelling a distance of $S$ and time taken to cover distance $S$ by sound. 

 $\text{Time taken by bullet to reach wall –}\\ {{T}_{1}}=\dfrac{S}{v} \\ \text{Time take by sound to reach back to Ajay will be-} \\ {{T}_{2}}=\dfrac{S}{V} \\ \text{Total time,} \\ \Rightarrow T={{T}_{1}}+{{T}_{2}} \\ \Rightarrow T =\dfrac{S}{v}+\dfrac{S}{V} \\ \therefore T =S\left( \dfrac{V+v}{V.v} \right)$

If a newly developed car works on some physics principles and its acceleration is given by the equation $a=-c{{v}^{{}^{3}/{}_{2}}}$. If ${{v}_{\circ }}$ is the initial velocity of the car and car owner wants to derive the relation between velocity and time from this equation to drive safely. This relation can be given as- 

 The first derivative of velocity function gives acceleration of a body and first derivative of displacement function gives velocity of a body.  

 $\text{The acceleration of car is given by,}\\ a=\dfrac{dv}{dt} \\ \Rightarrow \dfrac{dv}{dt}=-c{{v}^{3/2}} \\ \Rightarrow \dfrac{dv}{{{v}^{3/2}}}=-cdt \\ \text{Now integrating this function over v and t –} \\ \Rightarrow \int_{{{v}_{\circ }}}^{v}{\dfrac{dv}{{{v}^{3/2}}}}=\int_{0}^{t}{-c.dt} \\ \Rightarrow \left[ -\dfrac{2}{{v}^{1/2}} \right]_{{{v}_{\circ }}}^{v}=-ct \\ \Rightarrow -\dfrac{2}{\sqrt{v}}+\dfrac{2}{\sqrt{{{v}_{\circ }}}}=-ct \\ \Rightarrow \dfrac{2}{\sqrt{v}}=ct+\dfrac{2}{\sqrt{{{v}_{\circ }}}} \\ \therefore \dfrac{1}{v}={{\left( \dfrac{ct \sqrt{{{v}_{\circ }}}+2}{\sqrt{{{v}_{\circ }}}} \right)}^{2}}$